Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{-3y - 27}{y^3 + 4y^2 - 45y} \div \dfrac{y - 3}{y^3 + y^2 - 12y} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-3y - 27}{y^3 + 4y^2 - 45y} \times \dfrac{y^3 + y^2 - 12y}{y - 3} $ First factor out any common factors. $n = \dfrac{-3(y + 9)}{y(y^2 + 4y - 45)} \times \dfrac{y(y^2 + y - 12)}{y - 3} $ Then factor the quadratic expressions. $n = \dfrac {-3(y + 9)} {y(y + 9)(y - 5)} \times \dfrac {y(y - 3)(y + 4)} {y - 3} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-3(y + 9) \times y(y - 3)(y + 4) } { y(y + 9)(y - 5) \times (y - 3)} $ $n = \dfrac {-3y(y - 3)(y + 4)(y + 9)} {y(y + 9)(y - 5)(y - 3)} $ Notice that $(y + 9)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3y(y - 3)(y + 4)\cancel{(y + 9)}} {y\cancel{(y + 9)}(y - 5)(y - 3)} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $n = \dfrac {-3y\cancel{(y - 3)}(y + 4)\cancel{(y + 9)}} {y\cancel{(y + 9)}(y - 5)\cancel{(y - 3)}} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $n = \dfrac {-3y(y + 4)} {y(y - 5)} $ $ n = \dfrac{-3(y + 4)}{y - 5}; y \neq -9; y \neq 3 $